3.98.96 \(\int \frac {e^6 (4+2 x)+e^{\frac {2 e^5 x}{3}} (18 x+6 e^5 x^2)+e^{\frac {e^5 x}{3}} (e^3 (12+12 x)+e^8 (4 x+2 x^2))}{e^6} \, dx\)

Optimal. Leaf size=20 \[ \left (2+x+3 e^{-3+\frac {e^5 x}{3}} x\right )^2 \]

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Rubi [B]  time = 0.22, antiderivative size = 101, normalized size of antiderivative = 5.05, number of steps used = 21, number of rules used = 5, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} 6 e^{\frac {e^5 x}{3}-3} x^2+9 e^{\frac {2 e^5 x}{3}-6} x^2-36 e^{\frac {e^5 x}{3}-8} x+12 e^{\frac {e^5 x}{3}-3} x-36 e^{\frac {e^5 x}{3}-8}+(x+2)^2+36 e^{\frac {e^5 x}{3}-8} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^6*(4 + 2*x) + E^((2*E^5*x)/3)*(18*x + 6*E^5*x^2) + E^((E^5*x)/3)*(E^3*(12 + 12*x) + E^8*(4*x + 2*x^2)))
/E^6,x]

[Out]

-36*E^(-8 + (E^5*x)/3) - 36*E^(-8 + (E^5*x)/3)*x + 12*E^(-3 + (E^5*x)/3)*x + 6*E^(-3 + (E^5*x)/3)*x^2 + 9*E^(-
6 + (2*E^5*x)/3)*x^2 + 36*E^(-8 + (E^5*x)/3)*(1 + x) + (2 + x)^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (e^6 (4+2 x)+e^{\frac {2 e^5 x}{3}} \left (18 x+6 e^5 x^2\right )+e^{\frac {e^5 x}{3}} \left (e^3 (12+12 x)+e^8 \left (4 x+2 x^2\right )\right )\right ) \, dx}{e^6}\\ &=(2+x)^2+\frac {\int e^{\frac {2 e^5 x}{3}} \left (18 x+6 e^5 x^2\right ) \, dx}{e^6}+\frac {\int e^{\frac {e^5 x}{3}} \left (e^3 (12+12 x)+e^8 \left (4 x+2 x^2\right )\right ) \, dx}{e^6}\\ &=(2+x)^2+\frac {\int e^{\frac {2 e^5 x}{3}} x \left (18+6 e^5 x\right ) \, dx}{e^6}+\frac {\int \left (12 e^{3+\frac {e^5 x}{3}} (1+x)+2 e^{8+\frac {e^5 x}{3}} x (2+x)\right ) \, dx}{e^6}\\ &=(2+x)^2+\frac {\int \left (18 e^{\frac {2 e^5 x}{3}} x+6 e^{5+\frac {2 e^5 x}{3}} x^2\right ) \, dx}{e^6}+\frac {2 \int e^{8+\frac {e^5 x}{3}} x (2+x) \, dx}{e^6}+\frac {12 \int e^{3+\frac {e^5 x}{3}} (1+x) \, dx}{e^6}\\ &=36 e^{-8+\frac {e^5 x}{3}} (1+x)+(2+x)^2-\frac {36 \int e^{3+\frac {e^5 x}{3}} \, dx}{e^{11}}+\frac {2 \int \left (2 e^{8+\frac {e^5 x}{3}} x+e^{8+\frac {e^5 x}{3}} x^2\right ) \, dx}{e^6}+\frac {6 \int e^{5+\frac {2 e^5 x}{3}} x^2 \, dx}{e^6}+\frac {18 \int e^{\frac {2 e^5 x}{3}} x \, dx}{e^6}\\ &=-108 e^{-13+\frac {e^5 x}{3}}+27 e^{-11+\frac {2 e^5 x}{3}} x+9 e^{-6+\frac {2 e^5 x}{3}} x^2+36 e^{-8+\frac {e^5 x}{3}} (1+x)+(2+x)^2-\frac {18 \int e^{5+\frac {2 e^5 x}{3}} x \, dx}{e^{11}}-\frac {27 \int e^{\frac {2 e^5 x}{3}} \, dx}{e^{11}}+\frac {2 \int e^{8+\frac {e^5 x}{3}} x^2 \, dx}{e^6}+\frac {4 \int e^{8+\frac {e^5 x}{3}} x \, dx}{e^6}\\ &=-108 e^{-13+\frac {e^5 x}{3}}-\frac {81}{2} e^{-16+\frac {2 e^5 x}{3}}+12 e^{-3+\frac {e^5 x}{3}} x+6 e^{-3+\frac {e^5 x}{3}} x^2+9 e^{-6+\frac {2 e^5 x}{3}} x^2+36 e^{-8+\frac {e^5 x}{3}} (1+x)+(2+x)^2+\frac {27 \int e^{5+\frac {2 e^5 x}{3}} \, dx}{e^{16}}-\frac {12 \int e^{8+\frac {e^5 x}{3}} \, dx}{e^{11}}-\frac {12 \int e^{8+\frac {e^5 x}{3}} x \, dx}{e^{11}}\\ &=-108 e^{-13+\frac {e^5 x}{3}}-36 e^{-8+\frac {e^5 x}{3}}-36 e^{-8+\frac {e^5 x}{3}} x+12 e^{-3+\frac {e^5 x}{3}} x+6 e^{-3+\frac {e^5 x}{3}} x^2+9 e^{-6+\frac {2 e^5 x}{3}} x^2+36 e^{-8+\frac {e^5 x}{3}} (1+x)+(2+x)^2+\frac {36 \int e^{8+\frac {e^5 x}{3}} \, dx}{e^{16}}\\ &=-36 e^{-8+\frac {e^5 x}{3}}-36 e^{-8+\frac {e^5 x}{3}} x+12 e^{-3+\frac {e^5 x}{3}} x+6 e^{-3+\frac {e^5 x}{3}} x^2+9 e^{-6+\frac {2 e^5 x}{3}} x^2+36 e^{-8+\frac {e^5 x}{3}} (1+x)+(2+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.11, size = 58, normalized size = 2.90 \begin {gather*} 2 \left (2 x+\frac {x^2}{2}+\frac {9}{2} e^{-6+\frac {2 e^5 x}{3}} x^2+e^{\frac {e^5 x}{3}} \left (\frac {6 x}{e^3}+\frac {3 x^2}{e^3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^6*(4 + 2*x) + E^((2*E^5*x)/3)*(18*x + 6*E^5*x^2) + E^((E^5*x)/3)*(E^3*(12 + 12*x) + E^8*(4*x + 2*
x^2)))/E^6,x]

[Out]

2*(2*x + x^2/2 + (9*E^(-6 + (2*E^5*x)/3)*x^2)/2 + E^((E^5*x)/3)*((6*x)/E^3 + (3*x^2)/E^3))

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fricas [B]  time = 0.85, size = 42, normalized size = 2.10 \begin {gather*} {\left (9 \, x^{2} e^{\left (\frac {2}{3} \, x e^{5}\right )} + {\left (x^{2} + 4 \, x\right )} e^{6} + 6 \, {\left (x^{2} + 2 \, x\right )} e^{\left (\frac {1}{3} \, x e^{5} + 3\right )}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2*exp(5)+18*x)*exp(1/3*x*exp(5))^2+((2*x^2+4*x)*exp(3)*exp(5)+(12*x+12)*exp(3))*exp(1/3*x*exp(
5))+(2*x+4)*exp(3)^2)/exp(3)^2,x, algorithm="fricas")

[Out]

(9*x^2*e^(2/3*x*e^5) + (x^2 + 4*x)*e^6 + 6*(x^2 + 2*x)*e^(1/3*x*e^5 + 3))*e^(-6)

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giac [B]  time = 0.25, size = 107, normalized size = 5.35 \begin {gather*} \frac {1}{2} \, {\left (2 \, {\left (x^{2} + 4 \, x\right )} e^{6} + 9 \, {\left (2 \, x^{2} e^{10} - 6 \, x e^{5} + 9\right )} e^{\left (\frac {2}{3} \, x e^{5} - 10\right )} + 27 \, {\left (2 \, x e^{5} - 3\right )} e^{\left (\frac {2}{3} \, x e^{5} - 10\right )} + 12 \, {\left (x^{2} e^{10} + 2 \, x e^{10} - 6 \, x e^{5} - 6 \, e^{5} + 18\right )} e^{\left (\frac {1}{3} \, x e^{5} - 7\right )} + 72 \, {\left (x e^{5} + e^{5} - 3\right )} e^{\left (\frac {1}{3} \, x e^{5} - 7\right )}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2*exp(5)+18*x)*exp(1/3*x*exp(5))^2+((2*x^2+4*x)*exp(3)*exp(5)+(12*x+12)*exp(3))*exp(1/3*x*exp(
5))+(2*x+4)*exp(3)^2)/exp(3)^2,x, algorithm="giac")

[Out]

1/2*(2*(x^2 + 4*x)*e^6 + 9*(2*x^2*e^10 - 6*x*e^5 + 9)*e^(2/3*x*e^5 - 10) + 27*(2*x*e^5 - 3)*e^(2/3*x*e^5 - 10)
+ 12*(x^2*e^10 + 2*x*e^10 - 6*x*e^5 - 6*e^5 + 18)*e^(1/3*x*e^5 - 7) + 72*(x*e^5 + e^5 - 3)*e^(1/3*x*e^5 - 7))
*e^(-6)

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maple [B]  time = 0.06, size = 43, normalized size = 2.15

method	result	size
risch	\(x^{2}+4 x +9 x^{2} {\mathrm e}^{-6+\frac {2 x \,{\mathrm e}^{5}}{3}}+\left (6 x^{2} {\mathrm e}^{3}+12 x \,{\mathrm e}^{3}\right ) {\mathrm e}^{-6+\frac {x \,{\mathrm e}^{5}}{3}}\)	\(43\)
norman	\(\left (x^{2} {\mathrm e}^{3}+4 x \,{\mathrm e}^{3}+12 x \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}+6 x^{2} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}+9 x^{2} {\mathrm e}^{-3} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}}\right ) {\mathrm e}^{-3}\)	\(55\)
derivativedivides	\(3 \,{\mathrm e}^{-6} {\mathrm e}^{-5} \left ({\mathrm e}^{6} \left (\frac {x^{2} {\mathrm e}^{5}}{3}+\frac {4 x \,{\mathrm e}^{5}}{3}\right )+54 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}} x}{6}-\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}}}{4}\right )+54 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}} x^{2}}{18}-\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}} x}{6}+\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}}}{4}\right )+12 \,{\mathrm e}^{3} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}+12 \,{\mathrm e}^{3} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x \,{\mathrm e}^{5}}{3}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}\right )+36 \,{\mathrm e}^{3} {\mathrm e}^{-5} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x \,{\mathrm e}^{5}}{3}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}\right )+18 \,{\mathrm e}^{3} {\mathrm e}^{-5} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x^{2} {\mathrm e}^{10}}{9}-\frac {2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x \,{\mathrm e}^{5}}{3}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}\right )\right )\)	\(212\)
default	\({\mathrm e}^{-6} \left ({\mathrm e}^{6} \left (x^{2}+4 x \right )+162 \,{\mathrm e}^{-5} \left ({\mathrm e}^{-5} \left (\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}} x}{6}-\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}}}{4}\right )+{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}} x^{2}}{18}-\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}} x}{6}+\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{5}}{3}}}{4}\right )\right )+3 \,{\mathrm e}^{-5} \left (12 \,{\mathrm e}^{3} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}+12 \,{\mathrm e}^{3} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x \,{\mathrm e}^{5}}{3}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}\right )+36 \,{\mathrm e}^{3} {\mathrm e}^{-5} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x \,{\mathrm e}^{5}}{3}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}\right )+18 \,{\mathrm e}^{3} {\mathrm e}^{-5} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x^{2} {\mathrm e}^{10}}{9}-\frac {2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}} x \,{\mathrm e}^{5}}{3}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{3}}\right )\right )\right )\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2*exp(5)+18*x)*exp(1/3*x*exp(5))^2+((2*x^2+4*x)*exp(3)*exp(5)+(12*x+12)*exp(3))*exp(1/3*x*exp(5))+(2
*x+4)*exp(3)^2)/exp(3)^2,x,method=_RETURNVERBOSE)

[Out]

x^2+4*x+9*x^2*exp(-6+2/3*x*exp(5))+(6*x^2*exp(3)+12*x*exp(3))*exp(-6+1/3*x*exp(5))

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maxima [B]  time = 0.36, size = 45, normalized size = 2.25 \begin {gather*} {\left (9 \, x^{2} e^{\left (\frac {2}{3} \, x e^{5}\right )} + {\left (x^{2} + 4 \, x\right )} e^{6} + 6 \, {\left (x^{2} e^{3} + 2 \, x e^{3}\right )} e^{\left (\frac {1}{3} \, x e^{5}\right )}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2*exp(5)+18*x)*exp(1/3*x*exp(5))^2+((2*x^2+4*x)*exp(3)*exp(5)+(12*x+12)*exp(3))*exp(1/3*x*exp(
5))+(2*x+4)*exp(3)^2)/exp(3)^2,x, algorithm="maxima")

[Out]

(9*x^2*e^(2/3*x*e^5) + (x^2 + 4*x)*e^6 + 6*(x^2*e^3 + 2*x*e^3)*e^(1/3*x*e^5))*e^(-6)

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mupad [B]  time = 5.66, size = 33, normalized size = 1.65 \begin {gather*} x\,{\mathrm {e}}^{-6}\,\left ({\mathrm {e}}^3+3\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^5}{3}}\right )\,\left (4\,{\mathrm {e}}^3+x\,{\mathrm {e}}^3+3\,x\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^5}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-6)*(exp((x*exp(5))/3)*(exp(8)*(4*x + 2*x^2) + exp(3)*(12*x + 12)) + exp((2*x*exp(5))/3)*(18*x + 6*x^2
*exp(5)) + exp(6)*(2*x + 4)),x)

[Out]

x*exp(-6)*(exp(3) + 3*exp((x*exp(5))/3))*(4*exp(3) + x*exp(3) + 3*x*exp((x*exp(5))/3))

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sympy [B]  time = 0.19, size = 51, normalized size = 2.55 \begin {gather*} x^{2} + 4 x + \frac {9 x^{2} e^{3} e^{\frac {2 x e^{5}}{3}} + \left (6 x^{2} e^{6} + 12 x e^{6}\right ) e^{\frac {x e^{5}}{3}}}{e^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2*exp(5)+18*x)*exp(1/3*x*exp(5))**2+((2*x**2+4*x)*exp(3)*exp(5)+(12*x+12)*exp(3))*exp(1/3*x*e
xp(5))+(2*x+4)*exp(3)**2)/exp(3)**2,x)

[Out]

x**2 + 4*x + (9*x**2*exp(3)*exp(2*x*exp(5)/3) + (6*x**2*exp(6) + 12*x*exp(6))*exp(x*exp(5)/3))*exp(-9)

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